Question: Let $z_1$ and $z_2$ be the complex roots of $z^2 + az + b = 0,$ where $a$ and $b$ are complex numbers.  In the complex plane, 0, $z_1,$ and $z_2$ form the vertices of an equilateral triangle.  Find $\frac{a^2}{b}.$
Explanation: Let $z_2 = \omega z_1,$ where $\omega = e^{\pi i/3}.$  Then by Vieta's formulas,
\begin{align*}
-a &= z_1 + z_2 = (1 + \omega) z_1, \\
b &= z_1 z_2 = \omega z_1^2.
\end{align*}Hence,
\begin{align*}
\frac{a^2}{b} &= \frac{(1 + \omega)^2 z_1^2}{\omega z_1^2} \\
&= \frac{\omega^2 + 2 \omega + 1}{\omega} \\
&= \omega + 2 + \frac{1}{\omega} \\
&= e^{\pi i/3} + 2 + e^{-\pi i/3} \\
&= \frac{1}{2} + i \frac{\sqrt{3}}{2} + 2 + \frac{1}{2} - i \frac{\sqrt{3}}{2} \\
&= \boxed{3}.
\end{align*}